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A ball thrown in the air has a velocity of v(t)=82−32tft/sec at time t seconds. Find the total displacement of the ball between times t=1 second and t=6 seconds.

User Bocaxica
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Final answer:

The total displacement of the ball between times t=1 second and t=6 seconds is 180 ft.

Step-by-step explanation:

The total displacement of the ball between times t=1 second and t=6 seconds can be found by calculating the area under the velocity-time graph.

First, we need to find the area of the trapezoid formed by the velocity-time graph.

The formula to calculate the area of a trapezoid is A = (b1 + b2) * h / 2, where b1 and b2 are the lengths of the parallel sides and h is the height.

In this case, the parallel sides of the trapezoid are 82 ft/s and -46 ft/s, and the height is 5 seconds.

Therefore, the total displacement of the ball between times t=1 second and t=6 seconds is (82 + -46) * 5 / 2 = 180 ft.

User Shfx
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